## Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.2

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Question 1.

In the following expansions, find the term as stated :

(i) 5th term of (a + 2x^{3})^{17}

(ii) 9th term of \({ \left( \frac { x }{ y } -\frac { 3y }{ { x }^{ 2 } } \right) }^{ 17 }\)

(iii) 6th term of \({ \left( \frac { 2 }{ \sqrt { x } } -\frac { { x }^{ 2 } }{ 2 } \right) }^{ 9 }\)

Solution:

(i) 5th term of (a + 2x^{3})^{17
}We know that (r + 1 )^{th} term in the expansion of (a + b)^{n
}T_{r+1} = ^{n}C_{r} a^{n-r} b^{r
}Here a = ‘a’, b = 2x^{3}, n = 17

r + 1 = 5 ⇒ r = 4

Hence, 5th term = T_{5
⇒ }T_{5 }= T_{4+1
}_{⇒ }T_{5 }= ^{17}C_{4 }a^{17-4}(23x^{3})^{4
}_{⇒ }T_{5} = 2380 × a^{13} × 2^{4} × x^{12
}_{⇒ }T_{5} = 2380 × 16 × a^{13}x^{12
} _{⇒ }T_{5} = 38080 a^{13} x^{12
}Hence,

5th term = 38080 a^{13} x^{12} or ^{17}C_{4} a^{13} × 16x^{12
}

Question 2.

Find the coefficient of:

(i) x^{-7 }in the expansion of \({ { \left( ax-\frac { 1 }{ { bx }^{ 2 } } \right) }^{ 8 } }\)

(ii) x^{4} in the expansion of \({ { \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } \right) }^{ 15 } }\)

(iii) x^{6} in the expansion (a – bx^{2})^{10
}Solution:

(i) Coefficient of x^{-7} in the expansion of

In this term for coefficient of x^{-7} power of x should be

-7 = 8 – 3r = – 7

Hence, 8 – 3r = – 7

⇒ -3r = – 7 – 8 = – 15

r = 5

Hence, coefficient of x^{-7} in the expansion

(ii) Coefficient of x^{4} in the expansion of \({ { \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } \right) }^{ 15 } }\)

In this term, for coefficient of x^{4}, power of x should be 4

So, 60 – 7r = 4

⇒ -7r = 4 – 60 = – 56

⇒ r = 8

Hence, coefficient of x^{4} in the expansion = ^{15}C_{8} = 6435

(iii) Coefficient of x^{6} in the expansion of (a – b^{2})^{10
}

In this term, for coefficient of x^{6}, power of x should be 6.

So, 2r = 6 ⇒ r = 3

Hence, coefficient of x^{6} in the expansion

= (-1)^{3 10}C_{3} x a^{10-3} b^{3}

= – 1 × 120 × a^{7} b^{3}

= – 120 a^{7} b^{3}

Question 3.

In the following expansions find the term independent of x :

Solution:

(i) Term independent of x in the expansion of

for term independent of x

x^{12-3r }= x^{0
}⇒ 12- 3r = 0

⇒ -3r = -12

⇒ r = 4

Hence, term independent of x = (-1)^{4 12}C_{4
}= 1 × 495 = 495

(ii) Term independent of x in the expansion of

For the independent of x

Hence, term independent of x

= (-1)^{2 10}C_{2 }3^{2
}= 1 × 45 × 9 = 405

(iii) Term independent of x in the expansion of

\({ { \left( \sqrt { \frac { x }{ 3 } } +\frac { 3 }{ 2{ x }^{ 2 } } \right) }^{ 10 } }\)

Let (r + 1)^{th} term of expansion is constant i.e., independent of x.

(r + 1)^{th} term is constant i.e., power of x should be zero.

∴ 5- r/2 -2r = 0

(iv) Term independent of x in the expansion of

In this expansion for term independent of x, power of.v should be 0

20 – 4r = 0

⇒ – 4r = -20

⇒ r = 5

∴ Term independent of x in the expansion

= (-1)^{5 10}C_{5} =-252

Question 4.

In the following expansion, find the middle term :

Solution:

(i) Middle term in the expansion of \({ { \left( { \frac { x }{ 2 } }+2y \right) }^{ 6 } }\) In the expansion of \({ { \left( { \frac { x }{ 2 } }+2y \right) }^{ 6 } }\) middle term will be \({ T }_{ { { \left( { \frac { n }{ 2 } }+1 \right) } } }\) th term where x = 6 which is even number.

(ii) Middle term in the expansion of \({ { \left( { 3a }-\frac { { a }^{ 3 } }{ 6 } \right) }^{ 9 } }\)

Given expression is\({ { \left( { 3a }-\frac { { a }^{ 3 } }{ 6 } \right) }^{ 9 } }\)

Where n = 9 which is an odd number

∴ Middle term \(\frac { n+1 }{ 2 }+1 \)th term and \(\frac { n+1 }{ 2 } \)th term

i.e., Middle term is 5^{th} and 6^{th} term (T_{6} and T_{6})

Hence, two middle term are \(\frac { 189 }{ 8 } \) a^{17 }and – \(\frac { 21 }{ 16 } \) a^{19}

(iii) Middle term in the expansion of \({ { \left( x+\frac { 1 }{ x } \right) }^{ 2n } }\)

Given expansion is \({ { \left( x+\frac { 1 }{ x } \right) }^{ 2n } }\)

Where ‘n ’ = 2n, which is an even number

(iv) Middle term in the expansion of \({ { \left( 3x-\frac { 2 }{ { x }^{ 2 } } \right) }^{ 15 } }\)

Given expansion is \({ { \left( 3x-\frac { 2 }{ { x }^{ 2 } } \right) }^{ 15 } }\)

where n = 15, which is an odd number

∴ Middle term \(\frac { n+1 }{ 2 } \)th term and \(\frac { n+1 }{ 2 }+1 \)th term

i,e„ 8th and 9th term (T_{8} and T_{9})

Question 5.

Prove that if n is even then in expansion of (1 + x)^{n}, coefficient of middle term will be \(\frac { 1.3.5…..(n-1) }{ 2.4.6…..n } { 2 }^{ n }\)

If n is odd, then coefficient of both the middle term will be \(\frac { 1.3.5…..n }{ 2.4.6…..(n+1) } { 2 }^{ n }\)

Solution:

If n is even in (1 + x) term we take (1 + x)^{2n} in place of (1 + x)^{n} then ‘n’ = 2n is also even

Now, middle term = \(\left( \frac { 2n }{ 2 } +1 \right) \)th term

= (n + 1)^{th
}

Similarly, if n is odd, then we take (1 + x)^{2n+1} in place of (1 + x)^{n}, then ‘n’ = 2n + 1 is also odd. Now middle term are

Question 6.

If in the expansion of \({ \left( ax+\frac { 1 }{ bx } \right) }^{ 11 }\)

coefficient of x^{7} and x^{7} are equal then prove that ab – 1 – 0.

Solution:

(r + 1)^{th} term in the expansion of \({ \left( ax+\frac { 1 }{ bx } \right) }^{ 11 }\)

In this term, for coefficient of x^{7}, power of x should be 7.

Hence, 11 – 2r = 7

⇒ 2r = 11-7

⇒ 2r = 4,

⇒ r = 2

Coefficient of x^{7} in the expansion

Question 7.

In the expansion of (1 + y)^{n} if coefficients of 5^{th}, 6^{th} and 7^{th} terms are in A.P., then find the value of n.

Solution:

Question 8.

In Binomial expansion (x – a)^{n} second, third and fourth terms are 250, 720 and 1080 respectively. Find x, a and n.

Solution:

Second term in expansion of (x + a)^{n}

T_{2} = ^{n}C_{1} x^{n-1}.a

According to question,

Similarly, on dividing (3) by (2), we get

Question 9.

If in the expansion of (1+ a)^{n} coefficient of three consecutive terms are in ratio 1 : 7 ; 42, then find the value of n.

Solution:

Let in the expansion of (1 + x)^{n} three consecutive terms are (r – 1 )^{th} term, r^{th} term and (r + 1)^{th }term.

Here, (r – 1)^{th} term = ^{n}C_{r} – 2a^{r-2} and its coefficient ,

=^{n}C_{r} – 2

Similarly, coefficient of r^{th} term and (r + 1)^{th} term are ^{n}C_{r} -1 and ^{n}C_{r} respectively.

Now, given ratio of coefficient is 1 : 7 : 42

or 6r = n – r + 1

or n – 7r + 1 = 0 …….(2)

On solving (1) and (2), we get

n = 55

Question 10.

Find the positive value of m for which in the expansion of (1 + x)^{m} coefficient x^{2} is 6.

Solution:

From Binomial theorem

(1 + x)^{m} = ^{m}C_{0} + ^{m}C_{1}x + ^{m}C_{2 }x^{2} + ……

Then, coefficient of x^{2} in the expansion of (1 + x )^{m} = ^{m}C_{2
}